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The problem asks to find the value of \( k \) such that the function \( h(x) \), defined as:

\[
h(x) = 
\begin{cases} 
x^2 + 1 & \text{if } x < 2 \\
kx + 1 & \text{if } x \geq 2
\end{cases}
\]

is both continuous and differentiable at \( x = 2 \).

### Step 1: Continuity at \( x = 2 \)
For continuity at \( x = 2 \), the left-hand limit \( \lim_{x \to 2^-} h(x) \) must equal the right-hand limit \( \lim_{x \to 2^+} h(x) \), and both should equal \( h(2) \).

- For \( x < 2 \), \( h(x) = x^2 + 1 \). So:
  \[
  \lim_{x \to 2^-} h(x) = 2^2 + 1 = 4 + 1 = 5
  \]
  
- For \( x \geq 2 \), \( h(x) = kx + 1 \). So:
  \[
  \lim_{x \to 2^+} h(x) = 2k + 1
  \]

For continuity, \( \lim_{x \to 2^-} h(x) = \lim_{x \to 2^+} h(x) \), which gives:
\[
5 = 2k + 1
\]

Solving for \( k \):
\[
2k = 4 \quad \Rightarrow \quad k = 2
\]

### Step 2: Differentiability at \( x = 2 \)
For differentiability at \( x = 2 \), the derivatives from the left and right sides must be equal.

- The derivative of \( h(x) = x^2 + 1 \) (for \( x < 2 \)) is \( \frac{d}{dx}(x^2 + 1) = 2x \), so:
  \[
  \lim_{x \to 2^-} h'(x) = 2(2) = 4
  \]
  
- The derivative of \( h(x) = kx + 1 \) (for \( x \geq 2 \)) is \( \frac{d}{dx}(kx + 1) = k \), so:
  \[
  \lim_{x \to 2^+} h'(x) = k
  \]

For differentiability, the derivatives must match:
\[
4 = k
\]

### Conclusion:
There is **no value of \( k \)** that satisfies both the conditions of continuity and differentiability. Based on the calculations, \( k = 2 \) satisfies continuity but does not satisfy differentiability (because \( k \) would need to be 4 for differentiability). Thus, the answer is:

**(E) There

Let h be defined by the piecewise function h(x) = {x^2 + 1, for x < 2; kx + 1, for x ≥ 2}. For what value of k, if any, is h(x) both continuous and differentiable at x = 2?

Explore how to find the value of k that makes a piecewise function both continuous and differentiable at x = 2. This problem covers key concepts in calculus, including limits, continuity, and differentiability, suitable for advanced high school students.

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